最近剛學線段樹，做了些經典題目來練手
Just a Hook
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 24370 Accepted Submission(s): 12156

Problem Description
In the game of DotA, Pudge’s meat hook is actually the most horrible thing for most of the heroes. The hook is made up of several consecutive metallic sticks which are of the same length.

Now Pudge wants to do some operations on the hook.
Let us number the consecutive metallic sticks of the hook from 1 to N. For each operation, Pudge can change the consecutive metallic sticks, numbered from X to Y, into cupreous sticks, silver sticks or golden sticks.
The total value of the hook is calculated as the sum of values of N metallic sticks. More precisely, the value for each kind of stick is calculated as follows:
For each cupreous stick, the value is 1.
For each silver stick, the value is 2.
For each golden stick, the value is 3.
Pudge wants to know the total value of the hook after performing the operations.
You may consider the original hook is made up of cupreous sticks.

Input
The input consists of several test cases. The first line of the input is the number of the cases. There are no more than 10 cases.
For each case, the first line contains an integer N, 1<=N<=100,000, which is the number of the sticks of Pudge’s meat hook and the second line contains an integer Q, 0<=Q<=100,000, which is the number of the operations.
Next Q lines, each line contains three integers X, Y, 1<=X<=Y<=N, Z, 1<=Z<=3, which defines an operation: change the sticks numbered from X to Y into the metal kind Z, where Z=1 represents the cupreous kind, Z=2 represents the silver kind and Z=3 represents the golden kind.

Output
For each case, print a number in a line representing the total value of the hook after the operations. Use the format in the example.

Sample Input
1
10
2
1 5 2
5 9 3

Sample Output
Case 1: The total value of the hook is 24.

題目大意：Dota中的屠夫，有一個鉤子，鉤子分為n段，現在給出m個指令，將【X，Y】段鉤子變成Z種（Z分為三種：1為銅鉤，2為銀鉤，3為金鉤）初始鉤子都為Cu，m次操作后，求整個鉤的價值
每組樣例有t組數據
（N<=100000,1<=X<=Y<=N,1<=Z<=3）

維護一棵線段樹，區間修改，最后求一下全區間的和，唯一可以說的就是惰性標記是直接改變，并非不斷累積 因為所求是全區間的和，所以沒必要再額外寫一個區間求和的函數，直接在修改時下放標記，最后輸出這顆線段樹的根即可

代碼如下：

#include<iostream> #include<cstring> #include<algorithm> #include<cstdio> using namespace std; #define maxlen 100001 int value[maxlen<<2]={0},delta[maxlen<<2]={0}; int dota[maxlen]={0};void updata(int now) {value[now]=value[now<<1]+value[now<<1|1]; }void build(int l,int r,int now) {if (l==r){value[now]=dota[l];return;}int mid=(l+r)>>1;build(l,mid,now<<1);build(mid+1,r,now<<1|1);updata(now); }void pushdown(int now,int ln,int rn) {if (delta[now]!=0){delta[now<<1]=delta[now];delta[now<<1|1]=delta[now];value[now<<1]=delta[now]*ln;value[now<<1|1]=delta[now]*rn;delta[now]=0;} }void section_change(int L,int R,int l,int r,int now,int data) {if (L<=l && R>=r){value[now]=data*(r-l+1);delta[now]=data;return;}int mid=(l+r)>>1;pushdown(now,mid-l+1,r-mid);if (L<=mid)section_change(L,R,l,mid,now<<1,data);if (R>mid)section_change(L,R,mid+1,r,now<<1|1,data);updata(now); }int main() {int t;int n;int m;int time=1;scanf("%d",&t);while (true){if (t==0) break;scanf("%d",&n);for (int i=1; i<=n; i++)dota[i]=1;memset(delta,0,sizeof(delta));memset(value,0,sizeof(value));build(1,n,1);scanf("%d",&m);for (int i=1; i<=m; i++){int start,end,data;scanf("%d%d%d",&start,&end,&data);section_change(start,end,1,n,1,data);}printf("Case %d: The total value of the hook is %d.\n",time,value[1]);time++;t--;}return 0; }

轉載于:https://www.cnblogs.com/DaD3zZ-Beyonder/p/5346260.html

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